Ch 10Circles

Chapter 10: Circles

This chapter deals with the tangent to a circle — a line that touches the circle at exactly one point — and the two theorems (with proofs) that the board examination asks almost every year. Both proofs are prescribed for the exam.

1. Basic Definitions

• Circle: The collection of all points in a plane which are at a constant distance (the radius) from a fixed point (the centre).
• Secant: A line which intersects a circle in two distinct points.
• Tangent: A line which intersects (touches) the circle in exactly one point. The word comes from the Latin tangere, meaning "to touch".
• Point of contact: The common point of the tangent and the circle.
• The tangent is a special case of a secant when the two end points of its corresponding chord coincide.
• At any point on a circle there is one and only one tangent.

2. Number of Tangents from a Point

Length of a tangent: The length of the segment of the tangent from an external point P to the point of contact with the circle is called the length of the tangent from P.

3. Theorem 10.1 (Proof Required)

Proof (stepwise):

1. Given: A circle with centre O and a tangent XY to the circle at a point P. To prove: OP ⊥ XY.
2. Take any point Q on XY other than P and join OQ.
3. Since XY is a tangent, every point of XY except P lies outside the circle. So Q lies outside the circle, which gives OQ > radius, i.e. OQ > OP.
4. This holds for every point Q on XY other than P. Hence OP is the shortest distance from O to the line XY.
5. The shortest distance from a point to a line is the perpendicular distance. Therefore OP ⊥ XY. Hence proved.

Corollaries: (i) At any point on a circle there is only one tangent. (ii) The line perpendicular to the tangent at the point of contact passes through the centre. (iii) The perpendicular drawn from the centre to a tangent meets it at the point of contact.

4. Theorem 10.2 (Proof Required)

Proof (stepwise):

1. Given: A circle with centre O, an external point P, and two tangents PQ and PR to the circle from P, where Q and R are the points of contact. To prove: PQ = PR.
2. Join OP, OQ and OR.
3. By Theorem 10.1, ∠OQP = 90° and ∠ORP = 90°.
4. In right triangles OQP and ORP:
   OQ = OR (radii of the same circle)
   OP = OP (common hypotenuse)
5. Therefore ΔOQP ≅ ΔORP (RHS congruence rule).
6. Hence PQ = PR (CPCT). Hence proved.

Extra results from the same congruence (often asked): ∠OPQ = ∠OPR (OP bisects the angle between the tangents) and ∠POQ = ∠POR (OP bisects the angle between the radii). So the centre lies on the bisector of the angle between the two tangents.

5. Standard Results Proved Using the Theorems

• Quadrilateral circumscribing a circle: If a quadrilateral ABCD circumscribes (touches) a circle, then AB + CD = AD + BC. (Proof idea: write each side as a sum of two tangent lengths from its end vertices — e.g. AP = AS, BP = BQ, CR = CQ, DR = DS — and add.)
• Angle between two tangents: The angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre: ∠APB + ∠AOB = 180°. (In quadrilateral OAPB, the two angles at the points of contact are 90° each, and the angle sum is 360°.)
• Parallelogram circumscribing a circle is a rhombus. (Using AB + CD = AD + BC with AB = CD and AD = BC gives 2AB = 2BC, so all sides are equal.)
• Tangents at the ends of a diameter are parallel (both are perpendicular to the same diameter).
• Two concentric circles: A chord of the larger circle which touches the smaller circle is bisected at the point of contact, and its half-length = √(R² − r²).
• Incircle of a triangle: If the incircle touches BC, CA, AB at D, E, F, then BD = BF, CD = CE, AE = AF; with semi-perimeter s, BD = s − b, etc.

6. Key Computational Formula

7. Common Board Question Patterns and Methods

• Find tangent length / radius / distance: Draw the radius to the point of contact, mark the 90° angle, apply Pythagoras theorem in the right triangle.
• Angle chasing: Use (i) radius ⊥ tangent, (ii) ∠APB + ∠AOB = 180°, (iii) ΔOAP ≅ ΔOBP, (iv) isosceles triangle OAB (OA = OB). Example: if ∠APB = 80°, then ∠AOB = 100° and ∠POA = 50°.
• Proof questions: Theorem 10.1, Theorem 10.2, AB + CD = AD + BC, parallelogram circumscribing a circle is a rhombus, tangents at ends of a diameter are parallel, perpendicular at the point of contact passes through the centre.
• Incircle problems: Set tangent lengths from each vertex as x, y, z; form equations from the side lengths and solve.

Rationalised syllabus note: No topics have been deleted from this chapter. The full chapter, including both theorem proofs, is in the 2026-27 syllabus.